Prove that a closed subset of a complete normed vector space is complete.
My attempt,
Let S be a subset of $(V,\parallel . \parallel)$. Since S is closed, for all $x\in S$ there exists $\langle X_n \rangle_{n=1}^{\infty} \in S$ that converges to x. Therefore, $\forall \epsilon \gt 0 $ there exists $N \gt 0$ such that $\parallel x_n – x\parallel \lt \epsilon , n \ge N$. Therefore, every point in S corresponds to a bounded sequence in S, and since every bounded sequence has a convergent subsequence, every convergent sequence in S corresponds to a point in S.
Best Answer
Your proof is wrong. The fact that every convergent sequence has a limit does not imply that every Cauchy sequence has a limit. Furthermore it is not true that every bounded sequence has a convergent subsequence, this is true for $\mathbb R$, and some metric spaces, but not all.
For example take the unit sphere in an infinite dimensional vector space, and find a sequence of isolated points. The sequence is bounded, but it has no convergent subsequence.
The proof, however, is not very different. Suppose that $S$ is a closed subset of $V$, and let $\{x_n\}$ be a Cauchy sequence in $S$. Since $V$ is complete there is some $x\in V$ such that $x$ is the limit of $x_n$.
We want to show that $x\in S$. For every $\varepsilon>0$ there is some $N$ such that for all $n>N$, $\|x-x_n\|<\varepsilon$, and therefore for every $\varepsilon>0$ we can find some point $x'$ from $S$ such that $\|x-x'\|<\varepsilon$, and since $S$ is closed this means that $x\in S$, as wanted.